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习题4?2 1 1. 在下列各式等号右端的空白处填入适当的系数??使等式成立(例如??dx?d(4x?7): 4 (1) dx???d(ax); ?1 解dx???d(ax). a (2) dx? d(7x?3);?1 解dx? d(7x?3). 7 (3) xdx? d(x2); 解xdx? 1 d(x2). 2 (4) xdx? d(5x2);?1 解xdx? d(5x2). 10 (5)xdx? d(1?x2); 1 解 xdx? ? d(1?x2). 2 (6)x3dx? d(3x4?2);?1 解x3dx? d(3x4?2). 12 (7)e 2x dx? d(e2x); 解e 2x dx? 1 d(e2x). 2 (8)e2dx? d(1?e?x?xx?2); x?2 解 e2dx? ?2 d(1?e). 33 (9)sinxdx? d(cosx); 22323 解 sinxdx? ? d(cosx). 232 (10) 解 dx? d(5ln|x|); xdx1? d(5ln|x|). x5

(11) 解 dx? d(3?5ln|x|); xdx1? ? d(3?5ln|x|). x5dx (12)? d(arctan3x); 21?9xdx1 解 ? d(arctan3x). 231?9x (13)dx1?x2? d(1?arctanx); 解 dx1?x2? (?1) d(1?arctanx). (14)xdx1?x2xdx1?x2? d(1?x2). 解 ? (?1) d(1?x2). 2. 求下列不定积分(其中a, b, ?, ?均为常数): (1)?e5tdt; 解 ?e5tdt?15x15xed5x?e?C. 5?5 (2)?(3?2x)3dx; 解 ?(3?2x)3dx?? (3)? 解 1dx; 1?2x11(3?2x)3d(3?2x)??(3?2x)4?C. ?281111dx??d(1?2x)??ln|1?2x|?C. ?1?2x?21?2x2dx (4)?3; 2?3x 解 ?3?1131???(2?3x)3d(2?3x)???(2?3x)3?C??(2?3x)3?C. 33222?3xdx122

x(sinax?ebx(sinax?eb (5)? 解 )dx; 1x1)dx??sinaxd(ax)?b?ebd()??cosax?beb?C. abaxx? (6)?sinttsinttdt; 解 ?dt?2?sintdt??2cost?C. (7)?tan10x?sec2xdx;

解 ?tan10x?sec2xdx??tan10xdtanx? (8)? 解 dx; xlnxlnlnx1tan11x?C. 11dx11?dlnx??xlnxlnlnx?lnxlnlnx?lnlnxdlnlnx?ln|lnlnx|?C. x1?x2 (9)?tan1?x2?dx; 解 ?tan1?x?12x1?x2dx??tan1?xd1?x??22sin1?x2cos1?x2d1?x2 ??? (10)?cos1?x2dcos1?x2??ln|cos1?x2|?C. dx; sinxcosxdxsec2x1??dx??dtanx?ln|tanx|?C. 解 ?sinxcosxtanxtanx (11)? 解

1dx; x?xe?e?ex?e?x21ex1xxde?arctane?C. dx??2xdx??e?11?e2x (12)?xe?xdx;